If y = 5/2(e^(x/5) + e^(-x/5)), prove that y'' = y/25.

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If y = 5/2(e^(x/5) + e^(-x/5)), prove that y'' = y/25.

Mathematics
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I expanded the bracket: y = 5/2(e^(x/5)) + 5/2(e^(-x/5))
Then I took the first derivative:
y ' = 5x/10(e^(x/5)) - 5x/10(e^(-x/5))

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Now how should I go about with getting the second derivative? Should I factor out the 5x/10 and then use product rule?
\[y'=5/2[e ^{x/5}(1/5)+e^{-x/5}(-1/5)\]
\[y'=(5/2)(1/5)[e ^{x/5}-e ^{-x/5}]\]
Does it make a difference if we were to expand first with 5/2, and then take the first derivaitve?
not at all
Because that is what I did and then I took the dy/dx of each of the two terms.
but y did u multiply with x/5?
instead it is 1/5 the coefficient of x/5
No sorry I did the following: y = 5/2(e^{x/5}) + 5/2(e^{-x/5}) y '= 5/2(1/5)*(e^{x/5}) + 5/2(-1/5)(e^{-x/5})
correct
So then: y = 1/2(e^{x/5} + e^{-x/5})
y' u mean?
yes sorry, y'
y' = 1/2(e^{x/5} + e^{-x/5})
n there must b negative between to terms
coz 1/5 is the common factor
y' = 1/2(e^{x/5} - e^{-x/5}), right?
yes:)
So then for y'' I get..
\[y''=1/2[e ^{^{x/5}}(1/5)-e ^{-x/5}(-1/5)]\]
Yes I got that, then I expanded, and I got: y'' = 1/10(e^{x/5} + e^{-x/5})
\[y''=1/10[e ^{x/5}+e ^{-x/5}]\]
again no need to expand
\[e ^{x/5}+ e ^{-x/5}=2y/5\]
y''=y/25
How did you get e^x/5 + e^-x/5 = 2y/5?
OHHH
I think I know
got it?
I got it thank you so much. So when you are taking derivative of a function that has a constant which is multipled by a bracket that can be differentiated, you just take the derivative of the terms inside the bracket and multiply by the constant in front of it?
right:)
Ok awesome. Thank you for your help.
my pleasure

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